Boolean Semilattices - PDF Free Download

Boolean Semilattices Clifford Bergman Iowa State University June 2015

Motivating Construction Let G = G, be a groupoid (i.e., 1 binary operation) Form the complex algebra G + = Sb(G),,,,,, G X Y = { x y : x X, y Y } complex operation G + is an expansion of a Boolean algebra X = X = (normality) X (Y Z ) = (X Y ) (X Z ) and (Y Z ) X = (Y X) (Z X) (additivity) In fact, complete and atomic, and completely additive

Boolean Groupoids Definition A Boolean groupoid is an algebra B = B,,,,, 0, 1 such that B,,,, 0, 1 is a Boolean algebra x 0 0 x 0 x (y z) (x y) (x z) (y z) x (y x) (z x). BG = the variety of Boolean groupoids

Crash Course in Universal Algebra Algebra Structure with no relational parameters Variety Class of algebras closed under homomorphic image, subalgebra, and product Birkhoff s Theorem K is a variety iff it is axiomatized by a set of equations Equation Universal, atomic sentence Variety Generated by K V(K) is the smallest variety containing K. V(K) = Mod(Eq(K))

G a groupoid = G + BG Converse? Nope. G + ℵ 0 Is BG generated by complex algebras? Yes!

Can retrieve G from G + : a b = c in G {a} {b} = {c} in G + Note that {a}, {b}, {c} are atoms of G + Not all Boolean groupoids have atoms The product of two atoms might not be an atom We need a more general approach

Definition Let B be a Boolean groupoid. B + = A, θ in which A is the set of atoms of B and θ = { (x, y, z) A 3 : z x y } (a ternary relational structure) For a groupoid G, let θ = { (x, y, x y) }. Then G, θ = (G + ) +

Definition Let H = H, ψ be a ternary relational structure H + = Sb(H),,,,,, H in which X Y = { z H : ( x X)( y Y ) (x, y, z) ψ }. H a ternary relational structure = (H + ) + = H B a complete and atomic Boolean algebra = (B + ) + = B There is a duality between the categories of ternary relational structures and complete, atomic Boolean algebras

Canonical Extensions Theorem (Jónsson-Tarski, 1951) Let B be a Boolean groupoid. There is a complete, atomic Boolean groupoid B σ extending B such that p q = { a b : a, b B, a p, b q } for atoms p, q of B σ. Idea: (B σ ) + can serve as an approximation to a groupoid induced by B.

Lemma (Jipsen) Every Boolean groupoid can be embedded into P + for some partial groupoid P. (B σ ) + P G B B σ P + G +

Theorem Every Boolean groupoid lies in SH(G + ) for some groupoid G. Corollary BG = V { G + : G a groupoid }

V { G + : G a groupoid } is axiomatized by the identities that have to be true Will this hold for other classes of groupoids?

Properties Reflected in the Complex Algebra normality additivity monotonicity x y x z y z p q is linear if each variable appears exactly once in p and in q G p q G + p q x (y z) (x y) z, x y y x, x (x y) y p q is semilinear if p has no repeated variables and each variable of q occurs in p G p q G + p q x x 2, x y x

Some Generic Questions Let V be a variety of Boolean groupoids, K a class of groupoids or TRS. K + = { G + : G K } Is V = V(K + )? Is V = SP(K + )? ( V is representable by K ) Is V(K + ) finitely axiomatizable?

Some Nice Axiomatizations V { G + : G a commutative groupoid } axiomatized by x y y x V { G + : G an idempotent groupoid }: x x x V { G + : G a commutative, idempotent groupoid }: x y y x, x x x Together with axioms for Boolean algebras, normality, and additivity. (Jipsen) So all are finitely based. Are these varieties represented by a class of groupoids?

On the Other Hand... What about the associative law? Let Sg (CSg) be the variety of (commutative) semigroups. Theorem (Jipsen) Neither V(Sg + ) nor V(CSg + ) is finitely based.

On the Other, Other Hand... Lz = left-zero semigroups (x y x) V(Lz + ) is finitely based and represented by Lz. Rb = rectangular bands (x (y z) (x y) z, x x 2, x y z x z) V(Rb + ) is finitely based and represented by Rb.

Boolean Semilattices Let Sl denote the variety of semilattices, i.e. groupoids satisfying x (y z) (x y) z x y y x x x x. These are the identities of associativity, commutativity, and idempotence

Definition A Boolean semilattice is a Boolean groupoid satisfying The variety is denoted BSl. x (y z) (x y) z x y y x x x x

Easy to check that V(Sl + ) BSl. One wishes that these are equal. They are not. Example Let H = {a, b}. Define TRS on H with θ = { (a, a, a), (a, b, b), (b, a, b), (b, b, a), (b, b, b) }. H + BSl but Sl + x (y 1) x y while H + fails this identity

Open Questions 1 Is V(Sl + ) finitely based? Is the equational theory decidable? 2 Is either BSl or V(Sl + ) generated by its finite members? 3 Is there a finitely axiomatizable class K of TRS such that BSl = V(K + )?

Algebraic Theory of BSl Let B BSl, x B. x = x 1 Theorem yields a closure operator on B: x x = x and x y = x y x is closed if x = x

For a semilattice S and X S + X = X S = { y S : ( x X) y x } the downset generated by X.

Congruence Ideals Let θ Con(B). I = 0/θ = { x B : (0, x) θ } is an ideal of the Boolean algebra x I = x θ 0 = x θ 0 = 0 = x I Conversely I an ideal closed under implies I = 0/θ for some θ Con(B) Congruence ideals

Consequences Let B be a Boolean semilattice 1 Let a B. The smallest congruence ideal containing a is ( a]. 2 BSl has equationally definable principal congruences (EDPC) 3 B is subdirectly irreducible iff it has a smallest nonzero closed element 4 B is simple iff x > 0 = x = 1 Thus, for S Sl, S + is SI iff S has a least element S + is simple iff S is trivial 5 B subdirectly irreducible implies B σ subdirectly irreducible B simple implies B σ simple

Discriminator Algebras Let A be a set. The discriminator on A is { z if x = y d A (x, y, z) = x if x y. An algebra A is a discriminator algebra if d A is a term of A. Discriminator algebras are simple every subalgebra is a discriminator alg generate arithmetical varieties

Theorem Every simple BSl is a discriminator algebra, with term d(x, y, z) = ( x (x y) ) ( z (x y) )

Theorem The subvariety of BSl generated by all simple algebras is axiomatized, relative to BSl, by ( ( x) ) ( x) This is the largest discriminator subvariety of BSl. Problem: Is this variety representable by a nice family of TRS?

An Interesting Subvariety Tot = { H, H 3 : H any set }. Every member of Tot + is a discriminator algebra. V(Tot + ) is axiomatized (relative to BSl D ) by x y 1 x y.

Linear Semilattices Recall BSl x x x. What about IBSl = { B BSl : B x x x }? Let S Sl. S + x x x iff S is linearly ordered Proof: X X X = X is a subsemilattice So S has the property that every subset is a subsemilattice. Thus S is linearly ordered

Theorem Let LS denote the class of linearly ordered semilattices. Then IBSl = SP(LS + ).

In IBSl, x y = (x y) (y x) Thus IBSl is term-equivalent to its closure-reduct This variety of modal algebras is S 4,3

Consequences of EDPC For a class K, H ω (K) = { B/θ : B K, θ a compact congruence } If K BSl, then B/θ = B/(c] for a closed element c. A is splitting (relative to V) if subdirectly irreducible and there is a largest subvariety excluding A. Denoted V/A, defined by a single equation. Theorem (Blok-Pigozzi) If V has EDPC then every finite subdirectly irreducible algebra is splitting. In this case V/A = { B V : A / SH ω (B) }.

2 ℵ 0 subvarieties For n ω, a 1 a 2 a 3 a n Y n = Each Y + n is subdirectly irreducible, hence splitting. Easy to check: n m = Y + m / SH(Y + n ) Since Y n is finite, H(Y + n ) = H ω (Y + n )

Let T ω and V T = V { Y + n : n T }. Then m / T = Y + m / SH { Y + n : n T } = { Y + n : n T } BSl/Y + m Splitting, so BSl/Y + m is a variety. Therefore so Y + m / V T. V T = V { Y + n : n T } BSl/Y + m

m / T = Y + m / V T Thus S T = V S V T. Hence BSl has uncountably many subvarieties

Wrap-up The variety of Boolean semilattices is a very natural, finitely axiomatizable variety Has a rich arithmetic structure Numerous open problems on the border of all the components of BLAST Thanks for listening!